Polarisation & Malus's Law: JEE Main Guide

Polarisation contributes one to two questions per JEE Main session, making it a quick-win topic. The calculation demands are low — mostly Malus's law and Brewster's angle — but the conceptual understanding required is precise. Students who confuse unpolarised intensity with polarised intensity, or who forget the cos² rule, lose easy marks. This guide is your complete reference for every polarisation concept the exam uses.

Polarised vs Unpolarised Light

Unpolarised light has electric field oscillations in all directions perpendicular to the propagation direction. Polarised light has oscillations confined to one plane (plane of polarisation). When unpolarised light of intensity I₀ passes through an ideal polaroid (polariser), the transmitted intensity is I₀/2 regardless of the polariser orientation — because only the component of each random oscillation along the transmission axis passes, and averaging cos²θ over all angles from 0 to 2π gives ½. This I₀/2 result is the starting point for all Malus's law problems involving two polaroids.

Methods of polarisation: (1) by polaroids (selective absorption); (2) by reflection at Brewster's angle; (3) by scattering — sky is blue and partially polarised because scattering is proportional to 1/λ⁴, favoring shorter wavelengths; (4) by refraction (double refraction in crystals like calcite). JEE tests all four methods conceptually. Take a free wave optics mock to practise. See also our interference and diffraction guide.

Malus's Law

When plane-polarised light of intensity I₀ falls on an analyser (second polaroid) with its transmission axis making angle θ to the plane of polarisation: transmitted intensity I = I₀ cos²θ. This is Malus's law. If θ = 0°: I = I₀ (maximum). If θ = 90°: I = 0 (complete extinction). If θ = 45°: I = I₀/2. JEE question type 1: two polaroids with unpolarised incident light. After polariser: I₁ = I₀/2. After analyser at θ: I₂ = (I₀/2)cos²θ. JEE question type 2: three polaroids at 0°, 45°, 90°. After first: I₀/2. After second (45°): (I₀/2)cos²45° = I₀/4. After third (45° from second): (I₀/4)cos²45° = I₀/8. With two crossed polaroids, inserting a third at 45° transmits I₀/8 instead of zero — a classic conceptual question.

Brewster's Law

When light is reflected from a dielectric surface (glass, water) at Brewster's angle θ_B: tan θ_B = n (refractive index of the denser medium). The reflected ray is completely plane-polarised (electric field perpendicular to the plane of incidence only). The refracted ray is partially polarised. At Brewster's angle, reflected and refracted rays are perpendicular to each other (reflected + refracted = 90°). Numerical: for glass n = 1.5, θ_B = tan⁻¹(1.5) ≈ 56.3°. JEE asks you to find θ_B from n, or find n from θ_B, or identify whether a given angle produces complete polarisation.

Optical Activity and Applications

Optically active substances (sugar solution, quartz) rotate the plane of polarisation. Rotation angle α = [α]cl where [α] is specific rotation, c is concentration, l is path length. JEE uses this in straightforward proportionality questions: if the concentration doubles, the rotation doubles. Practical application: a polarimeter measures sugar concentration by measuring the rotation. The orientation of a polaroid cannot distinguish between clockwise and anticlockwise rotation — you need two polaroids (polariser + analyser) to quantify rotation. For the full wave optics strategy covering interference, see our ray and wave optics guide and for the broader optics chapter treatment see our modern physics guide.